Bài 6. Tính:
\(\eqalign{
& a)\mathop {\lim }\limits_{x \to + \infty } ({x^4} – {x^2} + x – 1) \cr
& b)\mathop {\lim }\limits_{x \to – \infty } ( – 2{x^3} + 3{x^2} – 5) \cr
& c)\mathop {\lim }\limits_{x \to – \infty } (\sqrt {{x^2} – 2x + 5}) \cr
& d)\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {{x^2} + 1} + x} \over {5 – 2x}} \cr} \)
:
\(\eqalign{
& a)\mathop {\lim }\limits_{x \to + \infty } ({x^4} – {x^2} + x – 1) = \mathop {\lim }\limits_{x \to + \infty } {x^4}\left( {1 – {1 \over {{x^2}}} + {1 \over {{x^3}}} – {1 \over {{x^4}}}} \right) = + \infty \cr
& b)\mathop {\lim }\limits_{x \to – \infty } ( – 2{x^3} + 3{x^2} – 5) = \mathop {\lim }\limits_{x \to – \infty } {x^3}\left( { – 2 + {1 \over x} – {5 \over {{x^2}}}} \right) = + \infty \cr
& c)\mathop {\lim }\limits_{x \to – \infty } (\sqrt {{x^2} – 2x + 5} ) = \mathop {\lim }\limits_{x \to – \infty } |x|\sqrt {1 – {2 \over x} + {5 \over {{x^2}}}} = + \infty \cr
& d)\mathop {\lim }\limits_{x \to + \infty } {{\sqrt {{x^2} + 1} + x} \over {5 – 2x}} = \mathop {\lim }\limits_{x \to + \infty } {{x\left( {\sqrt {1 + {1 \over {{x^2}}}} + 1} \right)} \over {5 – 2x}} = \mathop {\lim }\limits_{x \to + \infty } {{\left( {\sqrt {1 + {1 \over {{x^2}}}} + 1} \right)} \over {{5 \over x} – 2}} = – 1 \cr} \)
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